Proving that $\int f=\inf \left\lbrace\int g : g\text{ is a simple function such that } f\le g \right\rbrace $ For any bounded function $f:\mathbb{R}\to \mathbb{R}$, $$\int f=\inf \left\lbrace\int g : g\text{ is a simple function such that } f\le g \right\rbrace .$$
$f\le g$ implies that $\int f\le \inf \int g$.
Anyone can help me with finding a simple function $g\ge f$ with $\int g \le \int f + \epsilon$ ? Thank you.
 A: This is false in general. 
For instance, if $f$ is any integrable function which is positive everywhere (e.g., $x \mapsto e^{-x^2}$), then this is false. This is because any simple function which is greater than $f$ will have infinite integral.
The requirement for your statement to be true is that $f$ is bounded and nonzero only on a set of finite measure.
This is because we then have that $f \leq M\cdot \mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}$. Therefore, $M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-f \geq 0.$ Letting $g:=M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-f$, we can take a sequence $s_n$ of simple functions which converge to $g$ monotonically. Defining $t_n:=M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-s_n$, we have that $t_n \to M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-g=f$. 
By the monotone convergence theorem we have that $\int s_n \to M\mu(f^{-1}(\mathbb{R} \backslash \{0\}))-\int f$, and thus 
$$\int t_n=(M\mu(f^{-1}(\mathbb{R} \backslash \{0\}))-\int s_n ) \to (M\mu(f^{-1}(\mathbb{R} \backslash \{0\}))-(M\mu(f^{-1}(\mathbb{R} \backslash \{0\}))+\int f)=\int f.$$
By construction, $t_n:=M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-s_n \geq M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-g =f,$ and it is now easy to see that the result you state must hold.
